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(D) The given cycle is a Carnot cycle consisting of two isothermal and two adiabatic processes.The efficiency $\eta$ of a Carnot cycle is given by $\e

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$8$

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(D) The given cycle is a Carnot cycle consisting of two isothermal and two adiabatic processes.The efficiency $\eta$ of a Carnot cycle is given by $\eta = 1 - \frac{T_{L}}{T_{H}}$, where $T_{L}$ is the temperature of the cold reservoir and $T_{H}$ is the temperature of the hot reservoir.For an adiabatic process, $TV^{\gamma-1} = 	ext{constant}$.In the adiabatic compression step $(A 	o B)$, the gas is compressed from $V_{1}$ to $V_{B} = 1 \; m^{3}$ at temperature $T_{L}$ to $T_{H}$.Thus, $T_{L} V_{1}^{\gamma-1} = T_{H} V_{B}^{\gamma-1}$.$\frac{T_{L}}{T_{H}} = \left( \frac{V_{B}}{V_{1}} ight)^{\gamma-1} = \left( \frac{1}{V_{1}} ight)^{\gamma-1}$.Given $\eta = 3/4$, so $1 - \frac{T_{L}}{T_{H}} = 3/4$, which implies $\frac{T_{L}}{T_{H}} = 1/4$.For a monoatomic gas, the adiabatic index $\gamma = 5/3$, so $\gamma - 1 = 2/3$.Substituting these values: $\left( \frac{1}{V_{1}} ight)^{2/3} = 1/4$.Taking the reciprocal: $V_{1}^{2/3} = 4$.Raising both sides to the power of $3/2$: $V_{1} = 4^{3/2} = (2^{2})^{3/2} = 2^{3} = 8 \; m^{3}$.

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